As I said in my last long post, if you push on one end of a long rod, the other end will not move instantaneously. Using an idealized distributed spring mass model with no friction, damping, or dissipation, one gets the following results:
If the following is a long rod, and you push on the left end at x=0
(x=0)|||||||||||||||||||||||||(x=d)
The right end of the rod at x=d will first move at a time t = d/cs, where cs is the speed of sound in the rod.
The velocity of any point on the rod will be a function of the position on the rod, and the time, i.e.
velocity = v(t,x)
For 0<t<T, with T = d/c, (the length of the rod over the speed of sound in the rod)
v(x,t) will be given by
0 if t < x/c
v if t >= x/c
v is a constant depending on the exact characteristics of the rod and the applied force - it will be proportional to the applied force.
The plot of the above function would require three dimensions, x,t, and the dependent variable v(x,t). It represents, however, a "velocity wave" propagating to the right along the rod at the speed of sound.
For T < t < 2*T, v(x,t) will be
1*v if (t-T) < (d-x)/c
2*v if (t-T) >= (d-x)/c
This represents a wave moving to the left at the speed of sound. Like the previous wave, this wave is a plot of velocity vs time.
Note that the rod will compress during the time interval 0<t<T, and the rod will expand during the interval T<t<2T. The total change in the length of the rod will be
delta-L = v*d/c, where 'v' is the constant velocity previously mentioned, d is the length of the rod, and c is the speed of sound in the rod.
If delta-L/d is too large a quantity, the material will break, but usally delta-L will be very small.